Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $y = \dfrac{3t - 9}{4t^2 + 16t + 12} \div \dfrac{t - 2}{2t^2 + 2t - 12} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $y = \dfrac{3t - 9}{4t^2 + 16t + 12} \times \dfrac{2t^2 + 2t - 12}{t - 2} $ First factor out any common factors. $y = \dfrac{3(t - 3)}{4(t^2 + 4t + 3)} \times \dfrac{2(t^2 + t - 6)}{t - 2} $ Then factor the quadratic expressions. $y = \dfrac {3(t - 3)} {4(t + 3)(t + 1)} \times \dfrac {2(t + 3)(t - 2)} {t - 2} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac {3(t - 3) \times 2(t + 3)(t - 2) } { 4(t + 3)(t + 1) \times (t - 2)} $ $y = \dfrac {6(t + 3)(t - 2)(t - 3)} {4(t + 3)(t + 1)(t - 2)} $ Notice that $(t + 3)$ and $(t - 2)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac {6\cancel{(t + 3)}(t - 2)(t - 3)} {4\cancel{(t + 3)}(t + 1)(t - 2)} $ We are dividing by $t + 3$ , so $t + 3 \neq 0$ Therefore, $t \neq -3$ $y = \dfrac {6\cancel{(t + 3)}\cancel{(t - 2)}(t - 3)} {4\cancel{(t + 3)}(t + 1)\cancel{(t - 2)}} $ We are dividing by $t - 2$ , so $t - 2 \neq 0$ Therefore, $t \neq 2$ $y = \dfrac {6(t - 3)} {4(t + 1)} $ $ y = \dfrac{3(t - 3)}{2(t + 1)}; t \neq -3; t \neq 2 $